# Ex 15.2, 9

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex15.2, 9 Find the mean, variance and standard deviation using short-cut method Mean(π₯ Μ ) = A + (ββπππ¦π)/(ββππ) Γβ where A = assumed mean = 92.5 π¦_π=(π₯_π β π΄)/β h = class size = 75 β 70 = 5 Variance (π)2 = β2/N2 [πββπππ¦π2 β (ββπππ¦π)^2 ] = 52/602 [N Γ (254) β (6)2] = 25/3600 [60 Γ254 β36] = 25/3600 Γ 15204 = 105.58 β΄ Standard deviation (π) = βVariance = β105.58 = 10.27

Class 11

Important Question for exams Class 11

- Chapter 1 Class 11 Sets
- Chapter 2 Class 11 Relations and Functions
- Chapter 3 Class 11 Trigonometric Functions
- Chapter 4 Class 11 Mathematical Induction
- Chapter 5 Class 11 Complex Numbers
- Chapter 6 Class 11 Linear Inequalities
- Chapter 7 Class 11 Permutations and Combinations
- Chapter 8 Class 11 Binomial Theorem
- Chapter 9 Class 11 Sequences and Series
- Chapter 10 Class 11 Straight Lines
- Chapter 11 Class 11 Conic Sections
- Chapter 12 Class 11 Introduction to Three Dimensional Geometry
- Chapter 13 Class 11 Limits and Derivatives
- Chapter 14 Class 11 Mathematical Reasoning
- Chapter 15 Class 11 Statistics
- Chapter 16 Class 11 Probability

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.